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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf Complex geometry\\[15mm]
\small lecture 14: Dolbeault cohomology of a torus}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small Misha Verbitsky } 
\\[20mm]

{\tiny\bf HSE, room 306, 16:20,
\\[2mm] 
 November 7, 2020
}
\end{center}

\newpage

{\bf \blue Weight decomposition for $T^n$-action (reminder)}

\exercise
Consider the $n$-dimensional torus $T^n$ as a Lie group,
$T^n = U(1)^n$. {\bf \purple Prove that any finite-dimensional 
Hermitian representation of $T^n$ is a direct sum
of 1-dimensional representations,} with action of $T^n$ given by 
$\rho(t_1, ..., t_n)(x)= \exp(2\pi\1\sum_{i=1}^np_it_i) x$,
for some $p_1, ..., p_n\in \Z^n$, called {\bf \blue the
weights} of the 1-dimensional representation.

\definition
Let $V$ be a Hermitian space (possibly infinitely-dimensional)
equipped with an action of $T^n$,
and $V_\alpha\subset V$ weight $\alpha$ representations, $\alpha\in \Z^n$.
The direct sum $\bigoplus_{\alpha\in \Z^n} V_\alpha$ is called {\bf \blue
the weight decomposition} for $V$ if it is dense in $V$.

\theorem
Let $W$ be a Hermitian vector space. Then
{\bf \red the Fourier series provide the weight decomposition
on $L^2(T^n, W)$.}
\endproof

\theorem
Let $W$ be a Hermitian representation of $T^n$.
{\bf \red Then $W$ admits a weight decomposition
$V = \widehat{\bigoplus_{\alpha\in \Z^n} W_\alpha}$.}

\proof
We realize $W$ as a subrepresentation in
 $L^2(T^n, W)$, and use the Fourier series to
obtain the weight decomposition of  $L^2(T^n, W)$.
\endproof

\newpage

{\bf \blue Weight decomposition for $T^n$-action on differential forms
(reminder)}

\remark
Let $M$ be a manifold with the $T^n$-action,
and \[ \Lambda^*(M)= 
\hat\bigoplus_{\alpha \in \Z^n} \Lambda^*(M)_{p_1, ..., p_k}\]
be the weight decomposition on the differential forms.
Then the {\bf \red de Rham differential preserves each term
$\Lambda^*(M)_{p_1, ..., p_k}$.} Indeed, {\bf \purple $d$ commutes
with the action of the Lie algebra of $T^n$,
and $\Lambda^*(M)_{p_1, ..., p_k}$ are its
eigenspaces.}

\remark
Let $\alpha= \sum \alpha_{p_1, ..., p_k}$ be the weight decomposition.
{\bf \purple The forms $\alpha_{p_1, ..., p_k}$ 
are obtained by averaging
\[ 
e^{2\pi\1\sum_{i=1}^np_it_i}\alpha
= \Av_{T^n} e^{2\pi\1\sum_{i=1}^n-p_it_i}\alpha
\]
hence they are smooth.}


\newpage

{\bf \blue De Rham cohomology and $T^n$-action (reminder)}


\theorem
Let $M$ be a smooth manifold, and $T^n$ a torus
acting on $M$ by diffeomorphisms. Denote
by $\Lambda^*(M)^{T^n}$ the complex of 
$T^n$-invariant differential forms.
{\bf \red Then the natural embedding
$\Lambda^*(M)^{T^n}\hookrightarrow \Lambda^*(M)$
induces an isomorphism on de Rham cohomology.}

\pstep
Let $\alpha\in \Lambda^*(M)$ be a form
and $\alpha= \sum \alpha_{p_1, ..., p_n}$ its weight
decomposition, with $\alpha_{p_1, ..., p_n}\in \Lambda^*_{p_1, ..., p_n}(M)$ 
a form of weight $p_1, ... , p_n$.
 Since $T^n$-action
commutes with de Rham differential, 
these forms are closed when $\alpha$ is closed.

{\bf \green Step 2:}
Let $r_1, ..., r_n$ be the standard 
generators of the Lie algebra of $T^n$
rescaled in such a way that 
$\Lie_{r_k}(\exp(2\pi\1\sum_{i=1}^np_it_i))=\1 p_k$,
and $i_{r_k}:\; \Lambda^i(M)\arrow \Lambda^{i-1}(M)$
the convolution operator.
Since $\Lie_{r_k}= \{d, i_{r_k}\}$, we have
$p_k\alpha_{p_1, ..., p_n}= d (i_{r_k}\alpha_{p_1, ..., p_n})$
whenever $\alpha_{p_1, ..., p_n}$ is closed.
Therefore, {\bf \purple all terms in the weight 
decomposition $\alpha= \sum \alpha_{p_1, ..., p_n}$ 
are exact except $\alpha_{0,0, ...,0}$.}

{\bf \green Step 3:}
In the direct sum decomposition of the de Rham complex
\[
\Lambda^*(M)= \Lambda^*(M)^{T^n}\oplus 
\hat\bigoplus_{p_1,  ..., p_k\neq (0,0,..., 0)}\Lambda^*_{p_1, ..., p_k}(M)
\]
the second component has trivial cohomology, because
$\Lie_{r_k}$ is invertible on 
$\bigoplus_{p_k\neq 0}\Lambda^*_{p_1, ..., p_n}(M)$
{\bf \purple (deduce it from
$p_k\alpha_{p_1, ..., p_k}= d (i_{r_k}\alpha_{p_1, ..., p_k})$),} and\\
$\Lie_{r_k}(\text{closed form})$ is exact.
\endproof



\newpage

{\bf \blue Constant forms on a torus}

\remark In the proof above, the serie $\sum_{p_k\neq 0} \frac 1
        {p_k}i_{r_k}\alpha_{p_1, ..., p_k}$
converges, because $\sum_{p_k\neq 0}\alpha_{p_1, ...,
  p_k}$ converges, and satisfies $d\left( \sum_{p_k\neq 0} \frac 1
        {p_k}i_{r_k}\alpha_{p_1, ..., p_k}\right) 
=\sum_{p_k\neq 0}\alpha_{p_1, ...,  p_k}$ as shown. 

\definition
Let $T^n=(S^1)^n$ be a compact torus equipped with
a action on itself by shifts, and $\Lambda^*_\const(M)$.
the space of $T^n$-invariant forms on $T^n$.
These forms are called {\bf \blue constant
differential forms}. Clearly, {\bf \purple constant forms 
have constant coefficients in the usual
(flat) coordinates on the torus.}

\theorem
The natural embedding
$\Lambda^*_\const(T^n)\hookrightarrow \Lambda^*(T^n)$
{\bf \red induces an isomorphism $\Lambda^*_\const(T^n)=H^*(T^n)$.}

\proof The embedding
$\Lambda^*_\const(T^n)= \Lambda^*(T^n)^{T_n} \hookrightarrow \Lambda^*(T^n)$
induces an isomorphism on cohomology, however, all
constant
forms are closed, hence $H^*(\Lambda^*_\const(T^n), d)=\Lambda^*_\const(T^n)$.
\endproof

\newpage

{\bf \blue Holomorphic vector fields}

\definition
Let $(M,I)$ be a complex manifold, and
$X\in TM$ a real vector field. It is called
{\bf \blue holomorphic} if $\Lie_X(I)=0$,
that is, if the corresponding flow of
diffeomorphisms is holomorphic.

\claim
Let $(M,I)$ be a complex manifold, and
$X\in TM$ a holomorphic vector field.
{\bf \red Then $X^c:= I(X)$ is also holomorphic,
and commutes with $X$.}

\pstep
Assume that $X$ is non-zero at a given point $m\in M$
Solving the appropriate differential
equation in holomorphic coordinates, we obtain
a coordinate system $z_1, ..., z_n$ in a neighbourhood of
$m$ such that
$\Lie_X z_i= 0$ for $i>1$ and $\Lie_{z_1}=1$.
Let $x_i, y_i$ be the corresponding real 
coordinate system, wih $x_i =\Re z_i$ and
 $y_i =\Im z_i$. Then $X= \frac d{dx_1}$ and
$X^c= \frac d{dy_1}$.

{\bf \green Step 2:}
The conditions $\Lie_{X^c}(I)=0$ and
$[X^c,X]=0$ hold on a closed subset of $M$,
that is, they are true on the closure $C$
of the set of points where $X\neq 0$.
Outside of $C$, the vector field $X$ is
identically zero, hence these conditions
are also hold.
\endproof

\newpage

{\bf \blue Cartan's formula for Dolbeault differential}

\lemma
Let $X$ be a holomorphic vector field, and $X^c=I(X)$.
{\bf \red Then $\{d^c, i_X\}= - \Lie_{X^c}$.}

\proof Using $\{IdI^{-1}, i_X\}= I\{d,  I^{-1}i_XI\}I^{-1}$,
we obtain $\{d^c, i_X\}= - I \{d, i_{X^c}\}I^{-1}= I\Lie_{X^c}I^{-1}$.
However, $X^c$ is holomorphic, hence $I\Lie_{X^c}I^{-1}=
\Lie_{X^c}$.
\endproof


\proposition
Let $X$ be a holomorphic vector field, and $X^c=I(X)$.
{\bf \red Then $\{\bar\6, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})$.}

\proof
$\bar\6= \frac 1 2 (d +\1 d^c)$, hence
\[
\{\bar\6, i_X\}= \frac 1 2 \Lie_X + \1 \{d^c, i_X\}= \frac 1 2  
(\Lie_X - \1 \Lie_{X^c}).
\]
\endproof


\remark Let $M$ be a complex manifold
equipped with a holomorphic action of the torus 
$T^n$. Then the action of $T^n$ commutes with $d$ and $d^c$.
Therefore, the operators $d, d^c$ preserve 
the eigenspaces of the corresponding Lie algebra.
These eigenspaces are components of the weight decomposition.
This implies that {\bf \purple the Dolbeault differential
$\bar\6$ preserves the weight decomposition.}


\newpage

{\bf \blue Dolbeault cohomology of an elliptic curve}


\definition
{\bf \blue An elliptic curve} is a 1-dimensional
compact complex manifold $X:=\C/\Z^2$.

\remark
The additive group $\C$ acts on itself by parallel 
transforms, hence {\bf \red the 2-dimensional torus $T^2$ acts
on an elliptic curve by holomorphic diffeomorphisms.}

\definition
The $T^n$-invariant forms on $T^n$ are called
{\bf \blue constant}.

\definition
{\bf \blue Dolbeault cohomology} of a complex manifold
is $\frac{\ker \bar\6}{\im\bar\6}$.

\corollary 
{\bf \red Dolbeault cohomology of an elliptic curve $X$
are represented by the constant forms on $X$.}

{\bf \green Proof using the Hodge theory:}
Choose a $T^2$-invariant K\"ahler form on $X$.
We have already obtained an isomorphism between 
de Rham cohomology and the constant forms.
Since the constant forms are harmonic,
there are no other harmonic forms. Now,
$\Delta_{\bar\6}=\frac 1 2 \Delta_d$, hence\\
\phantom{x}
\ \ \ \ \ \ constant forms = $\bar\6$-harmonic forms = Dolbeault cohomology.
\endproof

In the next slide, {\bf \green
we give a proof which is independent from the
Hodge theory.}

\newpage

{\bf \blue Dolbeault cohomology of an elliptic curve (2)}


\proposition
Let $X$ be an elliptic curve, and 
$\Lambda^*(X)= \bigoplus_{\alpha \in \Z^2} \Lambda^*(X)_{p_1,p_2}$
its weight decomposition under the $T^2$-action.
Consider the space $T^2$-invariant forms
$\Lambda^*(X)^{T^2}=\Lambda^*(X)_{0,0}$.
{\bf \red Then the natural embedding $\Lambda^*(X)^{T^2}
\hookrightarrow \Lambda^*(X)$ induces an isomorphism
of Dolbeault cohomology.}

\proof
Let $\alpha\in \Lambda^*(X)_{p_1,p_2}$
be a $\bar\6$-closed form, with $(p, q)\neq (0,0)$.
Suppose, for example, that $p\neq 0$, and $X$
is the generator of the corresponding component
of the Lie algebra such that $\Lie_X\alpha=p\1\alpha$.
Since $X^c$ belongs to the same Lie algebra,
we have $\Lie_{X^c}(\alpha)= v\alpha$, where
$v\in \1\R$. Then 
\[
\frac{\1p+ v}{2} \alpha= 
\frac 1 2 (\Lie_X - \1 \Lie_{X^c})\alpha=\{ \bar\6, i_X\}\alpha=\bar\6 i_X\alpha,
\ \ \ (***)
\]
hence $\alpha$ is $\bar\6$-exact. This implies that {\bf \purple $\bar\6$ 
has no cohomology on 
\[ \bigoplus_{p_1, p_2\neq (0,0)}\Lambda^*(X)_{p_1,p_2}.\]}
\endproof

\newpage

{\bf \blue $\bar\6$-exact top forms on an elliptic curve}


\claim
Let $\eta\in \Lambda^n(T^n)$ be a top form on a torus,
and $\nu =\sum_{\alpha\in \Z^n} \nu_\alpha$
its weight decomposition. {\bf \red Then 
$\int_{T^n} \nu = \int_{T^n} \nu_0$,
where $\nu_0$ denotes the $T^n$-invariant
component.} Moreover, {\bf \red  whenever $\int_{T^n} \nu=0$,
the component $\nu_0$ also vanishes.}

\proof
Let $\nu$ be a top form on 
a compact manifold, equipped with an action of $S^1$,
and $\nu =\sum \nu_i$ its weight decomposition.
{\bf \purple Then $\int_M \nu_i=0$ for all $i\neq 0$.} Indeed, the $S^1$-action
multiplies $\nu_i$ by a non-zero number, but
the integral is invariant under the action of diffeomorphisms.
\endproof



\proposition
Let $\eta\in \Lambda^2(X)$ be a form on an elliptic curve
such that $\int_X\eta=0$. {\bf \red Then $\eta$ is $\bar\6$-exact.}

\proof Consider the weight decomposition 
$\eta =\sum_{\alpha\in \Z^2} \eta_\alpha$.
Since $\int_M \eta=0$, the (0,0)-component
vanishes, and by (***) the form 
$\eta$ is $\bar\6$-exact.
\endproof



\newpage

{\bf \blue Dolbeault cohomology of a disk}

\corollary
Let $K\subset \C$ be a compact subset, $K^0$ its interior,
and $\eta\in \Lambda^2(K^0)$ a top form smoothly extending to 
a neighbourhood of $K$.
{\bf \red Then $\eta$ is $\bar\6$-exact.}

\proof
Choosing an appropriate lattice $\Z^2\subset \C$,
we may assume that $K$ is a subset of an elliptic curve $X$.
Since $\eta$ extends to a neighbourhood of $K$, we can 
use partition of unity to extend it to a smooth form 
$\tilde \eta$ on $X$. Applying the weight decomposition
$\tilde \eta =\sum_{\alpha\in \Z^2} \eta_\alpha$, we obtain
that the form $\eta- \eta_{0,0}$ is $\bar\6$-exact.
However, the constant part 
$\eta_{0,0}=\const\cdot dz \wedge d\bar z= \const \cdot\bar\6(\bar z dz)$ is also
$\bar\6$-exact.
\endproof

\corollary
Let $K\subset \C$ be a compact subset, $K^0$ its interior,
and $\mu\in \Lambda^2(K^0)$ a (0,1)-form smoothly extending to 
a neighbourhood of $K$. 
{\bf \red Then $\mu$ is $\bar\6$-exact.}

\proof By the previous corollary,
$\mu\wedge dz$ is $\bar\6$-exact: there exists
a (1,0)-form $\phi$ such that
$\bar\6\phi =\mu\wedge dz$.
However, for any (1,0)-form
$\phi$ there exists a function $\psi$ such that
$\psi dz= \phi$, which gives
$\bar\6\psi =\mu$ because 
the map $\Lambda^{0,1}(X)\stackrel {\wedge dz}\arrow \Lambda^{1,1}(X)$
is an isomorphism which commutes with $\bar\6$.
\endproof

\newpage

{\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma}

\definition
{\bf \blue Polydisc} $D^n$ is a product of $n$ discs $D\subset \C$.

\theorem
{\bf \blue (Poincar\'e-Dolbeault-Grothendieck lemma)}\\
Let $\eta \in \Lambda^{p,q}(D^n)$, $q>0$, be a $\bar\6$-closed
form on a polydisc, smoothly extended to a neighbourhood
of its closure $\overline{D^n}\subset \C^n$. {\bf \red Then
$\eta$ is $\bar\6$-exact.}


{\bf \green We proved it for $n=1$.} 
Nextr lecture we prove it for all  $n$.




\end{document}





